Question

A block is released from rest on an fixed incline plane from a height 6m above the ground as shown in figure where coefficient of friction
between block and incline plane is 0.5. Choose the correct option(s) for motion of block from shown position to the bottom of incline :-​

Answer 1

Answer:

3.7

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Answer 2

The weight of the block is 5kg and the plane is inclined at an angle of 37°.

The weight of the block acts vertically downwards as

w = mg;

w = 5 × 9.81

w = 49.05 N

The normal reaction exterted by the plane of the block can calculated by resolving the weight component acting on the block.

Thus we get,

N = normal reaction = mg × cos Ф

where Ф = angle of incline

N = 49.05 × cos37

N = 39.17 N (acting perpendicular to the surface of incline)

Thus, resolving the weight component, the sinФ acts downwards along the slope of the incline and the friction force opposed it.

It has been given that α = coefficient of friction = 0.5

Thus friction force = α × N

F = 0.5 × 39.17

F = 19.585 N

While the weight component acting along the incline = mg × sinФ

G = 49.05 × sin37

G = 27.51 N

As G > F, the block moves downwards.

∴ G - F = ma;

where a = acceleration of the block

∴ 27.51 -  19.585 = 5 × a

∴ a = 1.585 m/s²

As the height of the slope and angle is given, you can find the run

∴ The run of slope = $\frac{height}{cos Ф}$

∴ run = 7.51 m